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2 years ago

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A supply capsule w/ crew approaches a spy satellite at a relative speed of 0.250 m/s so as to dock. The astronauts will then rep

air said spy satellite. The supply capsule has a mass of 4000 kg, and the spy satellite has a mass of 7500kg. a. Calculate the final velocity (after docking) by using the frame of reference in which the Spy satellite appears to be originally at rest. b. What is the loss of kinetic energy in this inelastic collision

1 answer:

WINSTONCH [101]2 years ago

7 0

The final velocity and the loss of kinetic energy is mathematically given as

vf = 0.087 m/s
L-K.E= 81.5 J

<h3>What are the final velocity and the loss of kinetic energy?</h3>

Question Parameters:

Generally, the equation for the conservation of momentum is mathematically given as

m1v1 = m2v2

(7500 + 4000) * vf = 4000 * 0.250

vf = 0.087 m/s

b)

Generally, the equation for the loss of kinetic energy is mathematically given as

loss in kinetic energy = initial kinetic energy - final kinetic energy

Therefore

L-K.E= 0.50 * 4000 * 0.25^2 - 0.50 * (4000 + 7500) * 0.087^2

L-K.E= 81.5 J

In conclusion, the final velocity and the loss of kinetic energy

vf = 0.087 m/s

L-K.E= 81.5 J

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Answer:

7664.06249 N/m

Explanation:

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timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

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(b) $r_{p}=$ 9.95 x 10⁻¹ m

(c) ratio = 1800

Explanation: (a) <u>Potential</u> <u>difference</u> is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

so

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

$qvB=\frac{mv^{2}}{r}$

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

(c) Radius of curvature, if it was a electron:

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

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